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Current Question (ID: 16599)

Question:
$\text{Three charges, each } +q, \text{ are placed at the corners of an equilateral triangle } ABC \text{ of sides } BC, AC, \text{ and } AB. D \text{ and } E \text{ are the mid-points of } BC \text{ and } CA. \text{ The work done in taking a charge } Q \text{ from } D \text{ to } E \text{ is:}$
Options:
  • 1. $\frac{3qQ}{4\pi\varepsilon_0 a}$
  • 2. $\frac{3qQ}{8\pi\varepsilon_0 a}$
  • 3. $\frac{qQ}{4\pi\varepsilon_0 a}$
  • 4. $\text{zero}$
Solution:
$\text{Hint: } W = q\Delta V$ $\text{Explanation: As the triangle } ABC \text{ is equilateral, and equal charges are placed at the vertexes of the triangle.}$ $\text{Also, points } D \text{ and } E \text{ are equidistant from all three charges.}$ $\text{Therefore, by using symmetry we can say that the potentials at } D \text{ and } E \text{ are equal. i.e., } V_D = V_E$ $\text{The work done in taking a charge } Q \text{ from } D \text{ to } E \text{ is given by:}$ $W_{D \rightarrow E} = Q(V_E - V_D) = 0$ $\text{Hence, option (4) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}